臺北市立南港高級中學 113 學年度第 1 次正式教師甄試
(配分: 1~11 題 每題 8 分, 第 12 題 12 分.)
解答:f(x)=(x+1)100=C1000+C1001x+C1002x2+⋯+C100100x100⇒f′(x)=100(x+1)99=C1001+2C1002x+3C1003x2+⋯+100C100100x99=C10099+2C10098x+3C10097x2+⋯+100C1000x99⇒f′(1)=100⋅299=C10099+2C10098+3C10097+⋯+100C1000解答:{R=(−π2,π2)R1=(−π2,−π3]R2=[−π3,−π6]R3=[−π6,0]R4=[0,π6]R5[π6,π3]R6=[π3,π2)⇒R=R1∪R2∪R3∪R4∪R5∪R6∀x∈R,∃θ∈Ri,1≤i≤6∋x=tan−1θ.For seven real numbers, there existstwo numbers {x=tan−1θ1y=tan−1θ2,θ1,θ2∈Ri,1≤i≤6⇒0≤|θ1−θ2|≤π6⇒0≤tan|θ1−θ2|≤tanπ6⇒0≤x−y1+xy≤1√3.QED
解答:
由於四種餐點在五天中都要吃到,所以有一種餐點會吃兩次。因此我們可以假設四種情形:
甲:點2次A,其餘各1次,即AABCD來排列。但A、A、B三者不相鄰,只能排成A○A○B,C與D只能排在○的位置。AAB有3種排法,CD有2種排法,共有3X2=6種排法;
乙:點2次B,此情形與甲相同,共有6種排法。
丙:點2次C,其它ABD各一次。CCABD共有5!2!=60情形,但需扣除CC相鄰或AB相鄰。CC相鄰共有4!=24種情形,AB相鄰共有4!2!×2=24種情形,CC相鄰且AB相鄰共有3!×2=12種情形。因此丙有60-24-24+12=24
丁:點2次D,此情形與丙相同,共有24種排法。
甲+乙+丙+丁=6+6+24+24=60

解答:{D(0,0,0)C(1,0,0)A(0,1,0)B(1,1,0)E(0,1,1)H(0,0,1)⇒M∈¯BE⇒M(1−t,1,t),0≤t≤1⇒{¯AM=√2t2−2t+1¯MH=√2t2−4t+3⇒¯AM+¯MH2≥√¯AM⋅¯MH⇒¯AM⋅¯MH=√4t4−12t3+16t2−10t+3=√f(t)f′(t)=0⇒2t−1√2t2−2t+1+2t−2√2t2−4t+3=02−12t2−2t+1=2−22t2−4t+3⇒2t2=1⇒t=1√2⇒¯AM+¯MH=√2−√2+√4−2√2⇒(¯AM+¯MH)2=2+√2⇒¯AM+¯MH=2+√2
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假設{¯AQ⊥¯CD¯AB⊥¯CPO為正方形BCDE中心點,由於tanθ=√1−a2a,因此假設{¯OQ=a¯AO=√1−a2⇒¯AQ=1⇒¯AC=√1+a2⇒¯AB=¯AC=√1+a2⇒△ABC面積=12¯AB⋅¯PC=12¯CD⋅¯QA⇒√1+a2⋅¯PC=2a⋅1⇒¯PC=2a√1+a2⇒cosα=2¯PC2−¯EC22¯PC2=1−¯EC22¯PC2=1−8a28a1+a2=−a2
解答:m=3n⇒3n∑k=1k23n3+k3=m∑k=1k219m3+k3=m∑k=11m(k/m)219+(k/m)3=∫10x219+x3dx=∫10/91/91/3udu(u=1/9+x3)=[13lnu]|10/91/9=13ln10
解答:{D(0,0,0)C(1,0,0)A(0,1,0)B(1,1,0)E(0,1,1)H(0,0,1)⇒M∈¯BE⇒M(1−t,1,t),0≤t≤1⇒{¯AM=√2t2−2t+1¯MH=√2t2−4t+3⇒¯AM+¯MH2≥√¯AM⋅¯MH⇒¯AM⋅¯MH=√4t4−12t3+16t2−10t+3=√f(t)f′(t)=0⇒2t−1√2t2−2t+1+2t−2√2t2−4t+3=02−12t2−2t+1=2−22t2−4t+3⇒2t2=1⇒t=1√2⇒¯AM+¯MH=√2−√2+√4−2√2⇒(¯AM+¯MH)2=2+√2⇒¯AM+¯MH=2+√2
解答:
假設△AIE的高為h,見上圖.又¯AD=¯DE⇒∠AED=∠DAE=2∠IAE=2θ△ADE三邊長為5,5,6⇒△ADE面積=√8(8−5)(8−5)(8−6)=12△ADE=△AID+△AIE=12⋅5h+12⋅6h=11h2=12⇒h=2411cos∠AED=cos2θ=62+52−5260=35⇒sinθ=√1−cos2θ2=√15¯AIsinθ=h⇒¯AI=hsinθ=2411√5
解答:
△APQ△ABC=12⇒¯APׯAQ¯ABׯAC=12⇒¯APׯAQ10×9=12⇒¯APׯAQ=45在△APQ中⇒cos∠A=¯AP2+¯AQ2−¯PQ22¯APׯAQ⇒38=¯AP2+¯AQ2−¯PQ22×45⇒¯AP2+¯AQ2−¯PQ2=1354⇒¯PQ=√¯AP2+¯AQ2−1354由於¯AP2+¯AQ22≥√¯AP2ׯAQ2=¯APׯAQ=45⇒¯AP2+¯AQ2≥90⇒¯PQ=√¯AP2+¯AQ2−1354≥√90−1354=√2254=152
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解答:{(a2+b2)(12+12)≥(a+b)2(b2+c2)(12+12)≥(b+c)2(c2+a2)(12+12)≥(c+a)2⇒{√a2+b2≥(a+b)/√2√b2+c2≥(b+c)/√2√c2+a2≥(c+a)/√2⇒√a2+b2+√b2+c2+√c2+a2≥2√2(a+b+c)=2√2=√2
解答:{an=√3an−1−bn−1bn=an−1+√3bn−1⇒[anbn]=[√3−11√3][an−1bn−1]A=[√3−11√3]=2[cos(π/6)−sin(π/6)sin(π/6)cos(π/6)]⇒An=2n[cos(nπ/6)−sin(nπ/6)sin(nπ/6)cos(nπ/6)]⇒[a18b18]=A18[a0b0]=218[cos(3π)−sin(3π)sin(3π)cos(3π)][a0b0]=218[−100−1][a0b0]=218[−a0−b0]⇒{a18=−218a0b18=−218b0⇒a18+b18=−218(a0+b0)=−219
解答:{an=√3an−1−bn−1bn=an−1+√3bn−1⇒[anbn]=[√3−11√3][an−1bn−1]A=[√3−11√3]=2[cos(π/6)−sin(π/6)sin(π/6)cos(π/6)]⇒An=2n[cos(nπ/6)−sin(nπ/6)sin(nπ/6)cos(nπ/6)]⇒[a18b18]=A18[a0b0]=218[cos(3π)−sin(3π)sin(3π)cos(3π)][a0b0]=218[−100−1][a0b0]=218[−a0−b0]⇒{a18=−218a0b18=−218b0⇒a18+b18=−218(a0+b0)=−219
Case I: x≥y⇒{P(x,y)至右邊界(x=1)的距離≥|x−y|P(x,y)至下邊界(y=0)的距離≥|x−y|⇒{1−x≥x−yy≥x−y⇒{2x−y≤12y≥xCase II: x≤y⇒{P(x,y)至左邊界(x=0)的距離≥|x−y|P(x,y)至上邊界(y=1)的距離≥|x−y|⇒{x≥y−x1−y≥y−x⇒{2x≥y2y−x≤1四條直線所圍菱形如上圖,其中{P=(y=2x)∩(2y−x=1)=(13,23)Q=(x=2y)∩(2x−y=1)=(23,13)因此{兩平行直線距離=1√5¯PB=√53⇒菱形面積=1√5×√53=13
解答:(1)轉換矩陣A=[1414014151414140150141414151401414151414141415]⇒A2=[1980740740740191007401980740740191007407401980740191007407407401980191001980198019801980625]⇒A2[00001]=[19100191001910019100625]⇒625(2)An−1[00001]=[(1−f(n))/4(1−f(n))/4(1−f(n))/4(1−f(n))/4f(n)]⇒An[00001]=[(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4f(n+1)]⇒[1414014151414140150141414151401414151414141415][(1−f(n))/4(1−f(n))/4(1−f(n))/4(1−f(n))/4f(n)]=[(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4f(n+1)]⇒f(n+1)=14(1−f(n))+15f(n)=−120f(n)+14⇒(x,y)=(−120,14)
解答:(1)轉換矩陣A=[1414014151414140150141414151401414151414141415]⇒A2=[1980740740740191007401980740740191007407401980740191007407407401980191001980198019801980625]⇒A2[00001]=[19100191001910019100625]⇒625(2)An−1[00001]=[(1−f(n))/4(1−f(n))/4(1−f(n))/4(1−f(n))/4f(n)]⇒An[00001]=[(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4f(n+1)]⇒[1414014151414140150141414151401414151414141415][(1−f(n))/4(1−f(n))/4(1−f(n))/4(1−f(n))/4f(n)]=[(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4(1−f(n+1))/4f(n+1)]⇒f(n+1)=14(1−f(n))+15f(n)=−120f(n)+14⇒(x,y)=(−120,14)
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